Find the equation for a tangent line without derivatives.

Hint: Here we can use the concept of infinitesimals. Infinitesimals simply represent the quantity which is very less in comparison to any finite quantity but is not zero.The slope of the tangent line is the instantaneous slope of the curve. Using these basic information we can solve the above given question.

Complete step by step solution:
Given a tangent line. Now we need to find the equation for a tangent line without using derivatives. So for that we can use the concept of infinitesimals. Now we know that the slope of a tangent line is simply the instantaneous slope of the given curve.Now if we just increase the argument of the given function by an infinitesimal amount and then divide it by the same infinitesimal amount we can say that it is the slope of the given function if we take only the finite part of numerical and discard other terms. Now to prove it let’s take an example and find the tangent to\[f\left( x \right)\;{\text{at}}\;x = 3\] and let the \[f\left( x \right) = {x^3} - 2{x^2} + 5x + 6\].

Let $\varepsilon > 0$be an infinitesimal value, so we can write:
\[\dfrac{{f\left( {3 + \varepsilon } \right) - f\left( 3 \right)}}{\varepsilon }...........................\left( i \right)\]
Now by the definition equation (i) would represent the slope. Now substituting all the values we can write:
\[\dfrac{{f\left( {3 + \varepsilon } \right) - f\left( 3 \right)}}{\varepsilon } = \dfrac{{\left( {{{\left( {3 + \varepsilon } \right)}^3} - 2{{\left( {3 + \varepsilon } \right)}^2} + 5\left( {3 + \varepsilon } \right) + 6} \right) - \left( {{{\left( 3 \right)}^3} - 2{{\left( 3 \right)}^2} + 5\left( 3 \right) + 6} \right)}}{\varepsilon } \\
\Rightarrow \dfrac{{f\left( {3 + \varepsilon } \right) - f\left( 3 \right)}}{\varepsilon }= \dfrac{{\left( {27 + 27\varepsilon + 9{\varepsilon ^2} + {\varepsilon ^3}} \right) - 2\left( {9 + 6\varepsilon + {\varepsilon ^2}} \right) + 15 + 5\varepsilon + 6 - 27 + 18 - 15 - 6}}{\varepsilon } \\
\Rightarrow \dfrac{{f\left( {3 + \varepsilon } \right) - f\left( 3 \right)}}{\varepsilon } = \dfrac{{\left( {27 + 27\varepsilon + 9{\varepsilon ^2} + {\varepsilon ^3}} \right) - 18 - 12\varepsilon - 2{\varepsilon ^2} + 15 + 5\varepsilon + 6 - 27 + 18 - 15 - 6}}{\varepsilon } \\
\Rightarrow \dfrac{{f\left( {3 + \varepsilon } \right) - f\left( 3 \right)}}{\varepsilon } = \dfrac{{20\varepsilon + 7{\varepsilon ^2} + {\varepsilon ^3}}}{\varepsilon } \\
\Rightarrow \dfrac{{f\left( {3 + \varepsilon } \right) - f\left( 3 \right)}}{\varepsilon } = 20 + 7\varepsilon + {\varepsilon ^2}...................................\left( {ii} \right) \\ \]
Now to get the slope of tangent from (iv) we have to consider only the finite numerical terms and discard all others, such that all we have remaining would be 20. Therefore the slope of the tangent would be 20, and the tangent point would be:
$\left( {3,f\left( 3 \right)} \right) = \left( {3,30} \right)$
Now we have got the slope $m$and a point $\left( {{x_1},{y_1}} \right) = \left( {3,30} \right)$. Such that we can write the equation as:
$m = \dfrac{{\left( {y - {y_1}} \right)}}{{\left( {x - {x_1}} \right)}} \\
\Rightarrow 20 = \dfrac{{\left( {y - 30} \right)}}{{\left( {x - 3} \right)}} \\
\Rightarrow 20\left( {x - 3} \right) = \left( {y - 30} \right) \\
\Rightarrow 20x - 60 = y - 30 \\
\therefore 20x - y = 30..............................\left( {iii} \right) \\ $
Therefore the equation of the tangent would be $20x - y = 30$.

Note: Infinitesimals in calculus has a wide range of applications and was introduced to solve the discrepancies in many proofs of calculus.
$m = \dfrac{{\left( {y - {y_1}} \right)}}{{\left( {x - {x_1}} \right)}}$
The above formula is to find the slope of a line and can also be used to find the equation of a line if any one point is given to us.